7/8y-32=3/4y+32

Solution for 7/8y-32=3/4y+32 equation:

7/8y-32=3/4y+32

We move all terms to the left:

7/8y-32-(3/4y+32)=0


Domain of the equation: 8y!=0

y!=0/8

y!=0

y∈R



Domain of the equation: 4y+32)!=0

y∈R


We get rid of parentheses

7/8y-3/4y-32-32=0

We calculate fractions


28y/32y^2+(-24y)/32y^2-32-32=0

We add all the numbers together, and all the variables

28y/32y^2+(-24y)/32y^2-64=0

We multiply all the terms by the denominator


28y+(-24y)-64*32y^2=0

Wy multiply elements

-2048y^2+28y+(-24y)=0

We get rid of parentheses

-2048y^2+28y-24y=0

We add all the numbers together, and all the variables

-2048y^2+4y=0

a = -2048; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-2048)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-2048}=\frac{-8}{-4096}
=1/512
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-2048}=\frac{0}{-4096}
=0
$